Area of the regionbounded by the curve `y = "cos" x` between `x = 0` and `x = pi` is

Options

A. `2" sq units"`
B. `4" sq units"`
C. `3" sq units"`
D. `1" sq unit"`

Explanation

To find the area of the region bounded by the curve \( y = \cos x \) between \( x = 0 \) and \( x = \pi \), we can follow these steps:

### Step 1: Understand the curve and the limits
The curve \( y = \cos x \) oscillates between 1 and -1. At \( x = 0 \), \( \cos(0) = 1 \); at \( x = \frac{\pi}{2} \), \( \cos(\frac{\pi}{2}) = 0 \); and at \( x = \pi \), \( \cos(\pi) = -1 \). We need to find the area between the curve and the x-axis from \( x = 0 \) to \( x = \pi \).

### Step 2: Set up the integral
Since the curve goes below the x-axis between \( x = \frac{\pi}{2} \) and \( x = \pi \), we need to split the integral into two parts:
1. From \( 0 \) to \( \frac{\pi}{2} \) where \( \cos x \) is positive.
2. From \( \frac{\pi}{2} \) to \( \pi \) where \( \cos x \) is negative.

Thus, the area \( A \) can be expressed as:
\[ A = \int_{0}^{\frac{\pi}{2}} \cos x \, dx + \left| \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx \right| \]

### Step 3: Calculate the first integral
Now, we calculate the first integral:
\[ \int_{0}^{\frac{\pi}{2}} \cos x \, dx \]
The integral of \( \cos x \) is \( \sin x \):
\[ = \left[ \sin x \right]_{0}^{\frac{\pi}{2}} = \sin\left(\frac{\pi}{2}\right) - \sin(0) = 1 - 0 = 1 \]

### Step 4: Calculate the second integral
Next, we calculate the second integral:
\[ \int_{\frac{\pi}{2}}^{\pi} \cos x \, dx \]
Again, using the integral of \( \cos x \):
\[ = \left[ \sin x \right]_{\frac{\pi}{2}}^{\pi} = \sin(\pi) - \sin\left(\frac{\pi}{2}\right) = 0 - 1 = -1 \]
Since we are interested in the area, we take the absolute value:
\[ \left| -1 \right| = 1 \]

### Step 5: Combine the areas
Now, we add the areas from both intervals:
\[ A = 1 + 1 = 2 \]

### Final Answer
Thus, the area of the region bounded by the curve \( y = \cos x \) between \( x = 0 \) and \( x = \pi \) is \( 2 \) square units.

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